It seems one of the trivial things to do, creating an attenuator with passive components.
If we want to attenuate the signal by 6dB, then what is more easy, than putting two equal value resistors
in series and form a voltage divider.
But that is that the truth? Equal value yes, but with what value?
Fig.1 shows a classic voltage divider used to attenuate a signal.
Let as consider now how the signal flows.
The signal comes from a source that has an output impedance and the lowest output impedance you will encounter in professional audio equipment is not smaller that 50Ω. So, the Johnson noise created is -135.2dBu, a low limit.
The maximum signal an OpAmp can handle is +22dBu, as a result the maximum dynamic range is:
135.2+22 = 155.2dB. The dynamic range of the ear is about 130dB.
Let us now examine the circuit with different resistor values. They cannot be low, because they will overload the previous circuit, causing distortion and reduced headroom and the cannot be too high because the induce large Johnson noise. So what is the optimum solution?
If we use 1kΩ resistors, then the Johnson noise floor becomes -125.2dB, because the effective resistor value for two 1k parallel resistor is 500Ω as seen from the output. The input impedance as seen from the source is 2kΩ.
We just raised noise by 10dB and probably the next components in the signal chain will further raise this noise more.
Now, the standard input impedance for a good audio equipment is at least 10kΩ. It means we will have to use 5k resistors, output impedance 2.5k and this produces a Johnson noise of -118.2dBu.
Of course in most, cases resistors near 100k are used because 10k input impedance is not high enough.
For a 100k input impedance we use two 50k resistors with Johnson noise -108.2. That is 27dB more noise than when the signal arrived with -135.2dBu from a 50Ω source.
If we have to use large resistors to create 100k input impedance and even higher, what is the solution to avoid excessive noise contamination?
The solution is to buffer the voltage divider circuit using a good low-noise OpAmp in front of it that can drive heavy loads as shown in Fig.2. OpAmps in general have very high input impedance.
If we use an OpAmp like the low-noise NE5532 we can achieve high input impedance and low noise contamination. The noise produced by NE5532 at its output is -119dBu and we can use two 500 Ω resistors that can be handled by NE5532 as a load. The Johnson noise from those resistors is -128.2dbU.
The output noise of the OpAmp is attenuated just like the normal signal and the result is the half of -119dBu = -125dBu, so the total noise is the sum of noises -128.2dBu and -125dBu, which is -123.3dBu.
Remember that the 6dB attenuator (voltage divider with equal value resistors) is the worst case, since it presents the largest output impedance.
Table.1 Johnson noise for different voltage dividers, at 25oC
and 22kHz bandwidth.
If we want to attenuate the signal by 6dB, then what is more easy, than putting two equal value resistors
in series and form a voltage divider.
But that is that the truth? Equal value yes, but with what value?
Fig.1 shows a classic voltage divider used to attenuate a signal.
Let as consider now how the signal flows.
The signal comes from a source that has an output impedance and the lowest output impedance you will encounter in professional audio equipment is not smaller that 50Ω. So, the Johnson noise created is -135.2dBu, a low limit.
The maximum signal an OpAmp can handle is +22dBu, as a result the maximum dynamic range is:
135.2+22 = 155.2dB. The dynamic range of the ear is about 130dB.
Let us now examine the circuit with different resistor values. They cannot be low, because they will overload the previous circuit, causing distortion and reduced headroom and the cannot be too high because the induce large Johnson noise. So what is the optimum solution?
If we use 1kΩ resistors, then the Johnson noise floor becomes -125.2dB, because the effective resistor value for two 1k parallel resistor is 500Ω as seen from the output. The input impedance as seen from the source is 2kΩ.
We just raised noise by 10dB and probably the next components in the signal chain will further raise this noise more.
Now, the standard input impedance for a good audio equipment is at least 10kΩ. It means we will have to use 5k resistors, output impedance 2.5k and this produces a Johnson noise of -118.2dBu.
Of course in most, cases resistors near 100k are used because 10k input impedance is not high enough.
For a 100k input impedance we use two 50k resistors with Johnson noise -108.2. That is 27dB more noise than when the signal arrived with -135.2dBu from a 50Ω source.
If we have to use large resistors to create 100k input impedance and even higher, what is the solution to avoid excessive noise contamination?
The solution is to buffer the voltage divider circuit using a good low-noise OpAmp in front of it that can drive heavy loads as shown in Fig.2. OpAmps in general have very high input impedance.
If we use an OpAmp like the low-noise NE5532 we can achieve high input impedance and low noise contamination. The noise produced by NE5532 at its output is -119dBu and we can use two 500 Ω resistors that can be handled by NE5532 as a load. The Johnson noise from those resistors is -128.2dbU.
The output noise of the OpAmp is attenuated just like the normal signal and the result is the half of -119dBu = -125dBu, so the total noise is the sum of noises -128.2dBu and -125dBu, which is -123.3dBu.
Remember that the 6dB attenuator (voltage divider with equal value resistors) is the worst case, since it presents the largest output impedance.
Table.1 Johnson noise for different voltage dividers, at 25oC
and 22kHz bandwidth.